You must add 45 mL of the 80 % alcohol to the 30 % alcohol to get a 35 % solution.
You can use a modified dilution formula to calculate the volume of 80 % alcohol
V1×C1 + V2×C2 = V3×C3
Let the volume of 80 % mixture 1 = <em>x</em> mL. Then the volume of the final 35 % mixture 3 = (405 + <em>x</em> ) mL
(<em>x</em> mL×80 % alc) + (405 mL×30 % alc) = (405 + <em>x</em>)mL × 35 % alc
80x + 12 150 = 14 175 + 35 x
45x = 2025
x = 2025/45 = 45
<u>Answer:</u> The number of formula units present in 2 moles of 
are 
<u>Explanation:</u>
Formula units are defined as the number of molecules or atoms present in 1 mole of a compound or element respectively.
According to mole concept:
1 mole of a compound contains 
number of formula units
Here, 2 represents the number of moles of
We are given:
Moles of (glucose) = 2 moles
Number of formula units of 
Hence, the number of formula units present in 2 moles of are
Hello there!
From my calculations the answers are:
D. Mass
B. Gravity
B. Gravity
Hope This Helps You!
Good Luck :)
With the given formula, we can calculate the amount of CO₂ using the balance equation but we first need the moles of CH₄
1) to find the moles of CH₄, we need to use the ideal gas formula (PV= nRT). if we solve for n, we solve for the moles of CH₄, and then we can convert to CO₂. Remember that the units put in this formula depending on the R value units. I remember 0.0821 which means pressure (P) has to be in atm, volume (V) in liters, the amount (n) in moles, and temperature (T) in kelvin.
PV= nRT
P= 1.00 atm
V= 32.0 Liters
n= ?
R= 0.0821 atm L/mol K
T= 25 C= 298 K
let plug the values into the formula.
(1.00 x 32.0 L)= n x 0.0821 x 298K
n= (1.00 x 32.0 L )/ (0.0821 x 298)= 1.31 moles CH₄
2) now let's convert the mole of CH₄ to moles to CO₂ using the balance equation
1.31 mol CH₄ (1 mol CO₂/ 1 mol CH₄)= 1.31 mol CO₂
3) Now let's convert from moles to grams using the molar mass of CO₂ (find the mass of each atom in the periodic table and add them)
molar mass CO₂= 12.00 + (2 x 16.0)= 44.0 g/mol
1.31 mol CO₂ ( 44.0 g/ 1 mol)= 57.6 g CO₂
Note: let me know if you any question.
Okay... what are the following
